Question 1

Form 5-digit numbers without repetition using digits {0, 1, 2, 3, 4, 5}. How many such numbers are divisible by 3? (The first digit cannot be 0.)

Accepted Answer

Introduction / Context:A number is divisible by 3 if the sum of its digits is a multiple of 3. We must count 5-digit, no-repetition numbers from the 6 digits {0,1,2,3,4,5} with a non-zero leading digit. Given Data / Assumptions: Available digits: 0–5. Distinct digits; 5-digit numbers; first digit ≠ 0. Sum of all six digits is 15, divisible by 3. Concept / Approach:Select which single digit is omitted; the remaining 5 digits define the digit-sum. If the total sum of chosen digits is divisible by 3, every permutation yields a number divisible by 3 (subject to the no-leading-zero rule). Step-by-Step Solution:Let S6 = 0+1+2+3+4+5 = 15 ≡ 0 (mod 3).Sum of chosen five digits ≡ −(omitted digit) (mod 3). To get 0 (mod 3), the omitted digit must be 0 or 3.Case 1 (omit 0): digits {1,2,3,4,5} ⇒ all 5! = 120 arrangements valid.Case 2 (omit 3): digits {0,1,2,4,5} ⇒ total 5! = 120 permutations; subtract leading-zero cases (0 as first): 4! = 24 ⇒ 96 valid.Total = 120 + 96 = 216. Verification / Alternative check:Residue-class reasoning (mod 3) matches the omitted-digit argument; direct enumeration agrees. Why Other Options Are Wrong:210/217 are near but incorrect; 122/192 do not follow from the modular analysis. Common Pitfalls:Forgetting to exclude leading-zero permutations in the case where 0 is present. Final Answer:216

Question 2

A new flag has six vertical stripes, each stripe colored using some or all of the colors {yellow, green, blue, red}. No two adjacent stripes may have the same color. In how many distinct stripe colorings can the flag be made?

Accepted Answer

Introduction / Context:This counts sequences of length 6 over a 4-color alphabet with an adjacency restriction that prohibits identical neighbors. Given Data / Assumptions: Colors available independently per stripe: 4 choices initially. No two consecutive stripes may have the same color. Concept / Approach:Use multiplication with a recurrence flavor: first stripe has 4 choices; each subsequent stripe has 3 choices (anything but its immediate neighbor’s color). Step-by-Step Solution:Stripe 1: 4 ways.Stripes 2–6: each 3 ways independently of earlier stripes beyond the immediate neighbor.Total = 4 × 3^5 = 4 × 243 = 972 = 12 × 81. Verification / Alternative check:A simple recurrence a(n) = 3 a(n−1) with a(1) = 4 yields a(6) = 4 × 3^5. Why Other Options Are Wrong:Other products do not equal 972 and do not match the 4·3^5 logic. Common Pitfalls:Accidentally allowing 4 choices at each step or forbidding more than just the immediate neighbor’s color. Final Answer:12 x 81

Question 3

An intelligence agency forms 2-digit codes from distinct digits 0–9, with the first digit non-zero. A handwritten code can be misread upside down if both digits are from {0,1,6,8,9} and the 180° rotation maps it to another valid 2-digit code (e.g., …

Accepted Answer

Introduction / Context:Under 180° rotation, the only digits that remain valid digits are {0,1,6,8,9} with mappings 0→0, 1→1, 6↔9, 8→8. A code is confusable if both digits come from this set and its rotated reading is a different valid 2-digit code (non-zero first digit, digits distinct). Given Data / Assumptions: Total 2-digit codes with distinct digits and non-zero first digit: 9 × 9 = 81. Rotation map f: {0,1,6,8,9} → {0,1,9,8,6} and order reverses. Concept / Approach:Count confusable codes and subtract from 81. Confusion requires both digits from {1,6,8,9} (not 0 for the second digit), so that the rotated code also has a non-zero leading digit and remains a valid 2-digit code with distinct digits. Step-by-Step Solution:Confusable codes = ordered pairs (x, y) with x ∈ {1,6,8,9}, y ∈ {1,6,8,9}, x ≠ y, because the rotated code is f(y)f(x) with f(y) ∈ {1,6,8,9} ≠ 0 and f(x) ≠ f(y).Count = 4 × 3 = 12.Therefore, non-confusable codes = 81 − 12 = 69. Verification / Alternative check:Cases with y = 0 yield rotated codes starting with 0 and hence are not considered confusing under the “valid code” criterion. Why Other Options Are Wrong:80/78 assume excluding very few cases; 71 is close but overcounts confusable cases; 81 ignores the confusion constraint. Common Pitfalls:Treating 0 as allowable as the rotated leading digit; or allowing equal digits (not permitted). Final Answer:69

Question 4

Using the digits {1, 2, 3, 4, 5} without repetition, form 5-digit numbers. How many such numbers have the units (ones) digit greater than the tens digit?

Accepted Answer

Introduction / Context:We count permutations with a simple inequality constraint involving only the last two positions (tens and units digits). Given Data / Assumptions: Digits are 1–5, used once each. 5-digit permutations (all digits used). Constraint: units digit > tens digit. Concept / Approach:Symmetry on ordered pairs: for any unordered pair {a, b} with a ≠ b, exactly one of the two orders satisfies units > tens. Step-by-Step Solution:Choose which two digits occupy the last two places: C(5, 2) = 10 unordered choices; for each chosen pair, exactly 1 ordering has units > tens.Arrange the first three positions with the remaining 3 digits: 3! = 6.Total = 10 × 6 = 60. Verification / Alternative check:Out of all 5P2 = 20 ordered pairs for the last two positions, half satisfy the inequality; 20/2 = 10 ways for the last two, times 6 ways for the first three gives 60. Why Other Options Are Wrong:48 (= 2 × 4!) ignores the exact half-ordering logic; 54/72 are ad hoc counts. Common Pitfalls:Double-counting when choosing ordered pairs first; working with unordered pairs avoids that. Final Answer:60

Question 5

Five boxes in a row (positions 1–5) are each filled with exactly one ball, either red (R) or blue (B). No two adjacent boxes may both contain blue balls. Balls of the same color are identical. How many different arrangements (strings over {R, B}) ar…

Accepted Answer

Introduction / Context:We are counting binary strings of length 5 with no consecutive B’s. This classic constraint yields a Fibonacci-type recurrence. Given Data / Assumptions: Length n = 5. Alphabet {R, B} with “no BB” adjacency. R and B are indistinguishable beyond color (identical balls). Concept / Approach:Let a(n) be the number of valid strings of length n. Appending R to any valid string of length n−1 is always allowed; appending B is allowed only to strings of length n−1 that end in R. This leads to a(n) = a(n−1) + a(n−2), a Fibonacci recurrence. Step-by-Step Solution:Base cases: a(1) = 2 (R, B); a(2) = 3 (RR, RB, BR).Then a(3) = 5; a(4) = 8; a(5) = 13.Therefore, there are 13 valid arrangements for 5 boxes. Verification / Alternative check:Direct enumeration confirms: 8 with at most two B’s spaced apart plus 5 with exactly two B’s non-adjacent. Why Other Options Are Wrong:8 and 10 undercount; 15 and 22 overcount by allowing adjacent B’s. Common Pitfalls:Assuming each position always has 2 independent choices (which ignores the adjacency ban) or forgetting that RB and BR are both valid. Final Answer:13

Question 6

Out of 7 consonants and 4 vowels, in how many different ways can words consisting of 3 consonants and 2 vowels be formed, assuming that the letters chosen are then arranged in all possible orders?

Accepted Answer

Introduction / Context: This question is from permutations and combinations and focuses on counting the number of words (arrangements of letters) formed by selecting consonants and vowels from given sets. It checks whether you can combine the ideas of selection and arrangement correctly. Given Data / Assumptions: Number of consonants available = 7. Number of vowels available = 4. Each word must contain exactly 3 consonants and 2 vowels. All selected letters are arranged in all possible different orders. All letters are distinct and can be used at most once in each word. Concept / Approach: The total number of words is: Number of ways to choose 3 consonants out of 7 = combination C(7,3). Number of ways to choose 2 vowels out of 4 = combination C(4,2). Number of ways to arrange the selected 5 letters = 5! permutations. The final answer is the product: Total words = C(7,3) * C(4,2) * 5!. Step-by-Step Solution: Step 1: Compute C(7,3) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35. Step 2: Compute C(4,2) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6. Step 3: Compute 5! = 5 * 4 * 3 * 2 * 1 = 120. Step 4: Multiply the three results: 35 * 6 * 120. Step 5: First 35 * 6 = 210. Step 6: Then 210 * 120 = 25200. Step 7: Hence, total number of words = 25200. Verification / Alternative check: We can verify the reasoning logically. There are 35 possible sets of 3 consonants, and for each such set there are 6 possible sets of 2 vowels. That gives 35 * 6 = 210 distinct letter groups. For each group of 5 letters, there are 120 permutations. Therefore, 210 groups * 120 permutations per group = 25200 distinct words, which is exactly the value calculated above. Why Other Options Are Wrong: 52000 and 24400: These are large numbers that may arise from incorrect multiplications or using 7P3 and 4P2 instead of combinations in the selection stage. 120: This is only 5!, the number of ways to arrange 5 letters, and ignores the many ways to choose the letters themselves. 8400: This might come from multiplying some but not all of the required factors correctly. None of these match the full combination and permutation reasoning. Common Pitfalls: A frequent error is to confuse combinations and permutations. Some students use permutations while selecting consonants and vowels, which leads to double counting. Another common issue is forgetting to include the final arrangement step after selection. Always separate the process into selection (using combinations) and arrangement (using factorials), and multiply the results in the correct order. Final Answer: The number of different words that can be formed is 25200.

Question 7

Using the digits 0, 1, 2, 3 and 4, how many integers greater than 999 but not greater than 4000 can be formed, if repetition of digits is allowed?

Accepted Answer

Introduction / Context: This question focuses on counting 4 digit integers under specific constraints. The digits that can be used are limited, repetition is allowed, and there is a bound on how large the number can be. It tests careful case analysis and understanding of place value constraints. Given Data / Assumptions: Allowed digits: 0, 1, 2, 3, 4. Integers must be greater than 999 and not greater than 4000. Repetition of digits is allowed. We are effectively dealing with 4 digit numbers between 1000 and 4000 inclusive. Concept / Approach: All required numbers have 4 digits. Let the digits be d1 d2 d3 d4. The leading digit d1 cannot be 0, or else the number would not be 4 digit. There is also an upper bound: the number must not exceed 4000. We split into cases based on d1: Case 1: d1 = 1, 2 or 3. Case 2: d1 = 4, but the number cannot exceed 4000. Step-by-Step Solution: Step 1: For 4 digit numbers using digits 0 to 4, d1 cannot be 0. Step 2: Consider Case 1: d1 can be 1, 2 or 3. That gives 3 choices for d1. Step 3: For each such choice, digits d2, d3 and d4 can each be any of the 5 digits 0, 1, 2, 3, 4 (repetition allowed). Step 4: So number of possibilities for Case 1 = 3 * 5 * 5 * 5 = 3 * 125 = 375. Step 5: Consider Case 2: d1 = 4. The maximum allowed number is 4000, so the remaining digits must be 000; any larger combination like 4100 would exceed 4000. Step 6: Therefore, the only valid number with leading digit 4 is 4000 itself. Step 7: Count for Case 2 = 1. Step 8: Total number of valid integers = 375 + 1 = 376. Verification / Alternative check: We can confirm the logic by understanding that all 4 digit numbers from 1000 to 3999 are automatically valid as long as their digits are restricted to 0 to 4, which is handled in Case 1 with 375 combinations. For numbers starting with 4, only 4000 qualifies because any other choice for the remaining digits would create a value larger than 4000. This reasoning aligns with the numeric counts computed earlier. Why Other Options Are Wrong: 375: This omits the extra valid number 4000 from Case 2. 500 and 673: These overcount and would require including invalid leading digits or numbers above 4000. 250: This undercounts and would likely result from miscounting possibilities or forgetting that repetition is allowed. Common Pitfalls: A very common mistake is to ignore the upper bound of 4000 and simply count all 4 digit numbers with allowed digits, or to forget that 0 cannot appear as the first digit. Another pitfall is failing to treat the leading digit 4 as a separate case and accidentally including many numbers from 4001 to 4444, which exceed the given limit. Systematic case analysis helps avoid these issues. Final Answer: The number of integers greater than 999 but not greater than 4000 that can be formed is 376.

Question 8

From a group of 6 men and 4 women, in how many different ways can a committee of 5 persons be formed so that at least 2 members of the committee are women?

Accepted Answer

Introduction / Context: This question is a standard combinations problem with an additional constraint. You need to form a committee of 5 members from a mix of men and women, with the condition that at least 2 women must be included. It tests the ability to handle combination formulas and to sum over valid cases that satisfy the constraint. Given Data / Assumptions: Total men available = 6. Total women available = 4. Committee size = 5 persons. At least 2 women must be included. Every person is distinct, and each combination of individuals counts as a different committee. Concept / Approach: We need to consider all valid distributions of women and men that satisfy the condition "at least 2 women." The possible cases for the committee of 5 are: Case 1: 2 women and 3 men. Case 2: 3 women and 2 men. Case 3: 4 women and 1 man. For each case, we use combinations: Number of ways = C(4, number of women) * C(6, number of men). Finally, we sum the results of all valid cases. Step-by-Step Solution: Step 1: Case 1 (2 women, 3 men) Ways to choose women = C(4,2). Ways to choose men = C(6,3). Total for Case 1 = C(4,2) * C(6,3) = 6 * 20 = 120. Step 2: Case 2 (3 women, 2 men) Ways to choose women = C(4,3) = 4. Ways to choose men = C(6,2) = 15. Total for Case 2 = 4 * 15 = 60. Step 3: Case 3 (4 women, 1 man) Ways to choose women = C(4,4) = 1. Ways to choose men = C(6,1) = 6. Total for Case 3 = 1 * 6 = 6. Step 4: Add all valid cases: 120 + 60 + 6 = 186. Step 5: Therefore, total number of committees with at least 2 women = 186. Verification / Alternative check: As a check, we can compute the total number of committees without restriction and subtract those with fewer than 2 women. Total committees from 10 people = C(10,5) = 252. Committees with 0 women: choose 5 men from 6 = C(6,5) = 6. Committees with 1 woman: choose 1 woman from 4 and 4 men from 6 = C(4,1) * C(6,4) = 4 * 15 = 60. Committees with fewer than 2 women = 6 + 60 = 66. Committees with at least 2 women = 252 - 66 = 186, which matches our earlier answer. Why Other Options Are Wrong: 196 and 190: These are close but result from small counting mistakes, such as miscomputing one of the combination terms. 200: A rounded guess that does not match any careful combination calculation. 176: This undercounts and likely omits some valid case or miscalculates C(6,3) or similar terms. Common Pitfalls: Students sometimes consider only one distribution, such as exactly 2 women, instead of all cases with at least 2 women. Another error is to forget that combinations, not permutations, should be used because order within the committee does not matter. Some also double count by mixing cases or forget to include the case with all 4 women. Listing the cases systematically and using correct combination formulas helps avoid these issues. Final Answer: The number of possible committees of 5 persons with at least 2 women is 186.

Question 9

If the letters of the word SACHIN are arranged in all possible ways and written in dictionary order, at which serial number does the word SACHIN appear in that list?

Accepted Answer

Introduction / Context: This question tests the idea of lexicographic or dictionary order of permutations of letters. Instead of listing every arrangement of the word SACHIN, which would be very time consuming, we use a systematic counting method based on factorials. Understanding this method is very important for many permutation and combination questions used in aptitude exams. Given Data / Assumptions: The word is SACHIN. The letters are S, A, C, H, I and N. All six letters are distinct. We assume standard dictionary order based on the English alphabet. Alphabetical order of the letters: A, C, H, I, N, S. Concept / Approach: The rank of a word in dictionary order is found by counting how many valid permutations come before it. For each position, we count how many smaller letters (in alphabetical order) could appear there, multiply that count by the number of permutations of the remaining positions, and add these contributions. Finally, we add 1 for the word itself, because the count of permutations before it does not include the word SACHIN itself. Step-by-Step Solution: Step 1: List letters in alphabetical order: A, C, H, I, N, S. Step 2: First letter is S. Letters smaller than S among {A, C, H, I, N, S} are A, C, H, I, N (5 letters). Step 3: For each such smaller letter at the first position, the remaining 5 letters can be arranged in 5! ways. So count = 5 * 5! = 5 * 120 = 600. Step 4: After fixing S at the first position, we move to the second letter A. Now remaining letters are A, C, H, I, N in alphabetical order, and A is the smallest among them. Step 5: There are no letters smaller than A among the remaining letters to place at the second position, so contribution here is 0. Step 6: Now fix SA and consider the third letter C. Remaining letters are C, H, I, N in alphabetical order. C is the smallest, so again contribution here is 0. Step 7: Fix SAC and look at H. Remaining letters are H, I, N; H is the smallest, so contribution again is 0. Step 8: Fix SACH and examine I. Remaining letters are I, N; I is the smaller, so contribution is 0. Step 9: Fix SACHI and then N is the only letter left, so we are at the word SACHIN. Step 10: Total permutations before SACHIN = 600. Therefore rank of SACHIN = 600 + 1 = 601. Verification / Alternative check: A quick check is to notice that S is the largest letter alphabetically. Hence all permutations starting with any of the other five letters (A, C, H, I, N) will come before any word starting with S. Since there are 5! words starting with each of those choices, the count of words before any S-starting word is exactly 5 * 5! = 600. Within the block of S-starting words, SACHIN is the first because the remaining letters A, C, H, I, N are already in alphabetical order. So the next rank after the first 600 permutations must be 601, which confirms our calculation. Why Other Options Are Wrong: 600 would be the count of words strictly before SACHIN, not the rank of SACHIN itself. 602 and 603 overshoot the correct rank by adding too much. 604 is even further away and has no direct combinatorial justification here. Only 601 matches the correct method and reasoning for lexicographic rank with six distinct letters. Common Pitfalls: A very common mistake is to forget to add 1 at the end, thereby reporting the number of words before SACHIN instead of its rank. Another typical error is to assume that contributions will come from later positions even when the current letter is already the smallest among the remaining letters. Some students also wrongly consider that dictionary order is based on the word SACHIN itself rather than the alphabetical order of individual letters, which leads to confusion. Carefully ordering the letters and counting smaller letters at each step avoids these issues. Final Answer: Thus, the word SACHIN appears at serial number 601 when all permutations are listed in dictionary order.

Question 10

At a party there are 12 people and each person shakes hands exactly once with every other person present. How many distinct handshakes take place in total?

Accepted Answer

Introduction / Context: This is a classic handshake problem that uses combinations. It checks whether you can recognize that each handshake involves a unique pair of people and then count how many such pairs exist in a group of 12 people. Problems like this often appear in aptitude tests to assess understanding of basic combinatorics rather than brute force counting. Given Data / Assumptions: Number of people at the party = 12. Each handshake occurs between exactly two different people. Any pair of people shakes hands at most once. No person shakes hands with themselves. Concept / Approach: A handshake is uniquely determined by choosing a pair of distinct people. Therefore, the total number of handshakes is the number of ways of selecting 2 people out of 12. This is a combination because the order of people in a handshake does not matter. The required count is given by the binomial coefficient nC2, where n is the number of people. Here n = 12, so we calculate 12C2. Step-by-Step Solution: Step 1: Recognize that each handshake corresponds to a pair of people. Step 2: Use the combination formula nC2 = n * (n - 1) / 2. Step 3: Substitute n = 12 to get 12C2 = 12 * 11 / 2. Step 4: Compute 12 * 11 = 132. Step 5: Divide by 2 to get 132 / 2 = 66. Step 6: Therefore, the total number of distinct handshakes is 66. Verification / Alternative check: Another way to think of the situation is to imagine each of the 12 people counting how many other people they can shake hands with. Each person can shake hands with 11 others, so if you multiply 12 * 11 you get 132. However, this counts every handshake twice (once from each person's perspective). To correct for this double counting, divide by 2, which again gives 132 / 2 = 66 handshakes. This matches the combination approach and confirms that the answer is consistent. Why Other Options Are Wrong: Values like 72, 76, 64 and 78 arise from incorrectly multiplying without dividing by 2, or from random guesses. For example, 12 * 6 = 72 is a common incorrect idea where someone assumes each person shakes hands with 6 people, which is not stated in the problem. None of these alternatives are equal to 12C2. Only 66 correctly reflects the number of unique pairs in a group of 12 people. Common Pitfalls: Students often forget to divide by 2 when using the approach of total handshakes counted from each person, leading to 132 instead of 66. Others might mistakenly treat the problem as permutations, incorrectly using nP2 rather than nC2. Some might even attempt to list handshakes manually and lose track or double count. Remember that combinations are used when order does not matter, which is exactly the case for handshakes. Final Answer: The total number of distinct handshakes that take place at the party is 66.

Question 11

In how many different ways can the letters of the word LEADING be arranged such that all the vowels in the word always come together as one block?

Accepted Answer

Introduction / Context: This problem checks your understanding of permutations under a grouping condition, where certain letters (the vowels) must always appear together. Instead of treating all letters as separate, we temporarily treat the group of vowels as one combined unit. This idea is very common in permutation and combination questions and helps to simplify complex arrangement problems. Given Data / Assumptions: The word is LEADING. Letters in LEADING are L, E, A, D, I, N, G. Vowels: E, A, I. Consonants: L, D, N, G. All seven letters are distinct, so there are no repetitions. Vowels must always stay together as a single contiguous block. Concept / Approach: To ensure that vowels stay together, we group E, A and I into one block. This block acts like a single super letter. Together with the consonants L, D, N and G, we then count the permutations of these objects. Finally, we multiply this by the number of internal permutations of the vowels within their block. This multiplication rule arises because for each arrangement of the blocks, there are several internal arrangements of vowels. Step-by-Step Solution: Step 1: Treat E, A and I as one block which we can call V. Step 2: Now we have the objects: V, L, D, N, G. That is 5 distinct objects to arrange. Step 3: The number of ways to arrange 5 distinct objects in a line is 5! = 120. Step 4: Inside the vowel block V, the three vowels E, A and I can be arranged in 3! = 6 ways. Step 5: For each arrangement of the 5 objects, there are 6 internal arrangements of the vowels. Step 6: Total required arrangements = 5! * 3! = 120 * 6 = 720. Verification / Alternative check: We can quickly confirm the logic by checking extremes. If there were no restriction, the total permutations of LEADING would be 7! = 5040. Requiring vowels to be together naturally reduces this count. The calculated answer 720 is significantly less than 5040, which makes sense. Also, another mental check is to think of the positions of the vowel block: as we have 5 objects including the block, the block can appear in any of 5 positions relative to the consonants, and for each such placement the three vowels can permute among themselves in 6 ways. It is consistent with the 5! * 3! calculation. Why Other Options Are Wrong: Values like 520, 700 and 750 do not follow the factorial based reasoning arising from the grouping method. They are often the result of guessing or partial multiplication mistakes. For example, mistakenly using 4! instead of 5!, or forgetting to account for the internal arrangement of vowels, would produce wrong totals. Only 720 correctly represents the number of arrangements obtained by multiplying 5! and 3! together. Common Pitfalls: A very common mistake is to forget that the vowels themselves can be arranged in different orders inside the block, leading to undercounting. Another mistake is to count the arrangement of the 5 objects incorrectly, or to accidentally include arrangements where vowels are not adjacent. Some students treat vowels as fixed in a particular order, which again reduces the total count incorrectly. Always remember to treat the vowel block as one object first, arrange all objects, and then expand the block internally. Final Answer: The number of different arrangements where all vowels in LEADING come together is 720.

Question 12

Three non collinear points A, B and C lie on the same plane. How many different straight lines can be drawn by joining these three points pairwise?

Accepted Answer

Introduction / Context: This question checks basic understanding of how many straight lines can be formed from a set of given points. When points are non collinear, no three points lie on the same straight line. Such questions are common in elementary geometry and combinatorics, and they focus on reasoning about pairs of points. Given Data / Assumptions: There are three points: A, B and C. The points are non collinear, meaning no single straight line passes through all three points. A straight line is uniquely determined by any two distinct points. No other hidden constraints are mentioned. Concept / Approach: Any straight line in a plane can be defined by choosing two distinct points and joining them. Therefore, the number of distinct lines that can be formed from a set of points is the number of distinct pairs of points. Since the three points are non collinear, each pair forms a different line. Hence, we only need to count the number of pairs of points using a simple combination formula or direct reasoning. Step-by-Step Solution: Step 1: List all the possible pairs of points: (A, B), (B, C) and (A, C). Step 2: Each pair corresponds to a unique line because no three points are collinear. Step 3: Count the number of such pairs. There are exactly 3 pairs. Step 4: Alternatively, use the combination formula nC2 where n = 3. Step 5: Compute 3C2 = 3 * 2 / 2 = 3. Step 6: Therefore, the number of distinct lines formed is 3. Verification / Alternative check: You can also sketch three non collinear points and manually draw lines between each pair. You will see one line between A and B, one between B and C, and one between A and C. There is no way to draw any additional line that passes through two of these points without coinciding with one of the already drawn lines. This visual method confirms that the correct count is 3 distinct lines. Why Other Options Are Wrong: The value 6 could arise from mistakenly counting each pair twice with reversed order (for example, AB and BA as different). The value 2 suggests missing one line segment altogether. The value 4 suggests adding a line that does not correspond to a pair of given points. Only 3 correctly counts the unique lines determined by pairs of the three points A, B and C. Common Pitfalls: Students sometimes confuse segments and lines, or they overcount by considering direction as important when it is not. Others might try to imagine additional lines passing through a single point, but by definition we only consider lines determined by at least two of the given points. Remember that for n distinct points with no three collinear, the number of distinct lines is always nC2, so for 3 points it must be 3. Final Answer: The number of different straight lines that can be drawn through the three non collinear points A, B and C is 3.

Question 13

In how many different ways can the letters of the word OPTICAL be arranged so that all the vowels in the word always appear together in every arrangement?

Accepted Answer

Introduction / Context: This is another grouping problem involving permutations with a condition on vowels. The word OPTICAL contains both vowels and consonants, and we are asked to count arrangements where all the vowels appear side by side. Such questions test your ability to restructure the problem by temporarily treating a group of letters as a single combined letter and then using factorials to count arrangements. Given Data / Assumptions: The word is OPTICAL. Letters are O, P, T, I, C, A, L. Vowels: O, I, A. Consonants: P, T, C, L. All seven letters are distinct. Vowels must always appear together as one contiguous block. Concept / Approach: We group the vowels O, I and A into a single super letter. Then we have this block plus the consonants P, T, C and L to arrange. So, we first count how many ways to arrange these 5 distinct objects. Next, we count how many ways the vowels can arrange among themselves inside the block. Using the multiplication principle, we combine these counts to reach the final number of valid arrangements. Step-by-Step Solution: Step 1: Treat O, I and A as one block V. Step 2: The objects to arrange are V, P, T, C and L. Thus we have 5 distinct objects. Step 3: The number of ways to arrange 5 distinct objects in a row is 5! = 120. Step 4: Inside block V, the vowels O, I and A can be arranged in 3! = 6 different ways. Step 5: For each arrangement of the 5 objects, there are 6 possible internal arrangements of the vowels. Step 6: Total number of required arrangements = 5! * 3! = 120 * 6 = 720. Verification / Alternative check: Without any restriction, all the letters of OPTICAL can be arranged in 7! = 5040 ways. The condition that vowels must appear together clearly restricts the arrangements, so the answer should be significantly smaller than 5040. We found 720 valid arrangements, which is exactly 5040 divided by 7. This is reasonable because the vowel block behaves like a single letter among seven possible positions, and the internal arrangements of vowels adjust the count appropriately. A quick sanity check of the arithmetic also confirms that 5! = 120 and 3! = 6 give 720 when multiplied. Why Other Options Are Wrong: The value 360 would arise if someone incorrectly used 4! instead of 5! or made a similar partial counting error. Values like 700 or 120 are not of the factorial form needed here, and 120 corresponds to only the outer arrangements of the vowel block and consonants, ignoring the internal arrangements of the vowels. Therefore they all underestimate the total number of valid permutations. Only 720 correctly combines both outer and inner arrangements. Common Pitfalls: Frequently, students forget to consider the internal permutations of the grouped vowels, which leads them to answer 5! instead of 5! * 3!. Another issue is mistakenly splitting the vowel block in the middle during counting, which violates the requirement that vowels stay together. Also, some learners incorrectly treat vowels as indistinguishable, even though O, I and A are different letters. Carefully separating outer block arrangements from inner vowel arrangements prevents these mistakes. Final Answer: The number of different arrangements of OPTICAL in which all vowels always appear together is 720.

Question 14

From a total of 7 men and 3 women, in how many different ways can a group consisting of exactly 5 men and 2 women be formed?

Accepted Answer

Introduction / Context: This question tests your understanding of combinations where you need to choose a specific number of men and women from separate pools. It is a typical selection problem where order does not matter. You are required to use the combination formula separately for men and women and then combine the results using the multiplication principle. Given Data / Assumptions: Total number of men = 7. Total number of women = 3. We must form a group with exactly 5 men and exactly 2 women. Order of people in the group does not matter. All individuals are distinct. Concept / Approach: We choose men and women independently. The number of ways to choose 5 men from 7 is given by 7C5, and the number of ways to choose 2 women from 3 is given by 3C2. Since each choice of 5 men can be combined with each choice of 2 women, we multiply these two combination values. This is a direct application of the rule of product in combinatorics. Step-by-Step Solution: Step 1: Compute the number of ways to choose 5 men from 7: 7C5. Step 2: Use symmetry: 7C5 = 7C2 = 7 * 6 / 2 = 21. Step 3: Compute the number of ways to choose 2 women from 3: 3C2 = 3 * 2 / 2 = 3. Step 4: Multiply the two results to get the total number of groups. Step 5: Total groups = 21 * 3 = 63. Step 6: Therefore, there are 63 different ways to form such a group. Verification / Alternative check: Another way to confirm the answer is to list possible compositions logically. There is no flexibility in the counts because the group must have 5 men and 2 women, which means all 3 women are eligible combinations but we only choose 2 at a time. Counting 7C5 directly with factorials as 7! / (5! * 2!) and 3C2 as 3! / (2! * 1!) will still produce 21 and 3 respectively. Multiplying gives 63 again, so the result is consistent under different calculation approaches. Why Other Options Are Wrong: Values like 135, 125 and 64 do not match the product 7C5 * 3C2. For example, 7C3 * 3C2 would give 35 * 3 = 105, which still does not appear in the options. The choice 210 might arise from incorrectly choosing all 7 men or mixing counts in the wrong way. Only 63 is directly supported by correct combination formulas for each gender group. Common Pitfalls: Common mistakes include using permutations instead of combinations, forgetting that the order of people in the group does not matter, or accidentally choosing the wrong number of men or women. Some students may also incorrectly add combination counts instead of multiplying them, which is incorrect because each combination of men can be paired with each combination of women. Always treat independent selections with the multiplication rule and use nCr for counting choices where order is irrelevant. Final Answer: The required group can be formed in 63 different ways.

Question 15

A college has 10 basketball players. In how many different ways can a playing team of 5 members and a separate captain be selected from these 10 players?

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Introduction / Context: This problem involves selecting a team and then designating a captain from among the selected players. It combines combinations and a simple additional choice. It is a standard example of hierarchical selections, where you pick a subset and then pick a special role from within that subset. Such questions are common in aptitude exams and help to clarify the difference between selecting people and assigning roles. Given Data / Assumptions: Total number of basketball players = 10. We need to select a team of 5 members. One of the 5 selected players must be chosen as captain. All 10 players are distinct. Only one captain is chosen, and every team has exactly one captain. Concept / Approach: First, we choose 5 players out of 10 to form the team, where order does not matter. This is a combination 10C5. After the team is chosen, we then pick one of the 5 team members to be the captain. This is an additional factor of 5. Since these two stages are independent, we multiply the counts. This is an application of the rule of product in combinatorics: total ways = 10C5 * 5. Step-by-Step Solution: Step 1: Compute the number of ways to choose 5 players from 10: 10C5. Step 2: 10C5 = 10! / (5! * 5!) = 252. Step 3: For each chosen group of 5 players, any one of the 5 can be chosen as captain. Step 4: Number of ways to choose the captain from the 5 is 5. Step 5: Multiply the two counts: total selections = 10C5 * 5 = 252 * 5. Step 6: Calculate 252 * 5 = 1260. Step 7: Therefore, there are 1260 different ways to form the team and select the captain. Verification / Alternative check: You can reverse the steps and first pick the captain and then the remaining 4 players. First, select the captain from the 10 players in 10 ways. After that, choose 4 more players from the remaining 9 to complete the team of 5. The count is 10 * 9C4. Now 9C4 = 126, so total = 10 * 126 = 1260, which matches the earlier method. The equality 10C5 * 5 = 10 * 9C4 is a nice consistency check and confirms our answer. Why Other Options Are Wrong: Values like 1400, 1250 or 1600 are not supported by correct combinatorial reasoning. They might come from using permutations directly or from arithmetic errors such as mistakenly computing 10P5 or mixing up factorial values. Only 1260 arises from both valid methods described above and correctly respects the hierarchy of selecting a team and then a captain. Common Pitfalls: A common error is to treat the problem as if we are only selecting a captain and four other players without recognizing that the captain must come from within the team. Another mistake is using permutations instead of combinations when forming the team, which overcounts arrangements where the same set of players appears in different orders. Some students also forget the final multiplication by 5 when selecting the captain. Always separate selections (combinations) from role assignments (additional choices) clearly. Final Answer: The number of ways to select a 5 member team and a captain from 10 players is 1260.

Question 16

How many different arrangements of all the letters of the word COMMITTEE are possible in which the four vowels in the word do not all appear together as a single block?

Accepted Answer

Introduction / Context: This problem involves permutations of a word with repeated letters, under a restriction on the placement of vowels. The word COMMITTEE contains repeated letters M, T and E, which means we must use permutations of multiset formulae. We first count all possible arrangements and then subtract the arrangements where all four vowels are together. This is a standard use of the complement method in permutation and combination problems. Given Data / Assumptions: Word: COMMITTEE. Letters: C, O, M, M, I, T, T, E, E. Vowels: O, I, E, E (four vowels with E repeated twice). Consonants: C, M, M, T, T. We consider all 9 letters in each arrangement. Condition: the four vowels must not all appear together as one single block. Concept / Approach: The total number of distinct permutations of COMMITTEE is found by dividing 9! by factorials of the counts of repeated letters. To enforce the condition that vowels do not all sit together, we compute two values: total arrangements and arrangements where all vowels are grouped as a single block. The required count is total minus the grouped vowel count. For the grouped case, we glue the four vowels into one super block and then treat the multiset of this block with the consonants, taking care of repeated consonants and internal permutations of vowels. Step-by-Step Solution: Step 1: Count total permutations of COMMITTEE. Step 2: There are 9 letters with repetitions: M appears twice, T appears twice, E appears twice. Step 3: Total arrangements = 9! / (2! * 2! * 2!). Step 4: Compute 9! = 362880 and 2! * 2! * 2! = 8. So total arrangements = 362880 / 8 = 45360. Step 5: Now count arrangements where all vowels O, I, E, E form one block. Step 6: Treat the vowel block V as one item. Together with consonants C, M, M, T, T we have 6 items: V, C, M, M, T, T. Step 7: Among these 6 items, M is repeated twice and T is repeated twice. Step 8: Arrangements of these 6 items = 6! / (2! * 2!) = 720 / 4 = 180. Step 9: Inside the vowel block V, the letters are O, I, E, E where E is repeated twice. Step 10: Internal arrangements of vowels = 4! / 2! = 24 / 2 = 12. Step 11: Total arrangements with all vowels together = 180 * 12 = 2160. Step 12: Required arrangements with vowels not all together = total - grouped = 45360 - 2160 = 43200. Verification / Alternative check: A reasonable check is to compare magnitudes. The number of grouped arrangements (2160) is relatively small compared to the total 45360, which makes sense because enforcing all four vowels to be together is a strict restriction. Subtracting 2160 leaves 43200, which is still large but slightly less than the full total. Also, computing with a calculator or careful manual arithmetic confirms 6! / (2! * 2!) = 180 and 4! / 2! = 12, and their product is 2160. Thus the arithmetic of the complement method holds correctly. Why Other Options Are Wrong: The option 45360 corresponds to the total number of permutations without any restriction, which ignores the requirement that vowels are not all together. The value 216 refers to a much smaller count and does not arise naturally from the factorials used here. The value 1260 is another incorrect mixture of factorial terms. Only 43200 equals 45360 minus 2160 and correctly represents the number of arrangements where the four vowels do not form one single block. Common Pitfalls: Common mistakes include forgetting to account for repeated letters, which leads students to use 9! instead of 9! divided by the repetition factorials. Others mistakenly count arrangements where the vowels are together instead of not together. Some also misinterpret the condition and try to forbid any two vowels from being adjacent rather than all four. Using the complement method and carefully respecting repeated letters avoids these errors. Always pay attention to the words together and not together in such questions. Final Answer: The number of different arrangements of COMMITTEE in which the four vowels do not all appear together is 43200.

Question 17

How many different 4 letter words, with or without meaning, can be formed from the letters of the word LOGARITHMS if no letter is repeated beyond what is available?

Accepted Answer

Introduction / Context: This question checks your ability to form permutations of a fixed length from a larger set of distinct letters. The word LOGARITHMS consists of 10 different letters. We are asked to form 4 letter sequences (words) using these letters without repetition. It is a direct application of permutations of distinct objects taken r at a time, written as nPr. Given Data / Assumptions: Word: LOGARITHMS. Letters: L, O, G, A, R, I, T, H, M, S. Total distinct letters available = 10. We need to form 4 letter words. No letter may be repeated within a single word. Order of letters within the word matters because different orders give different words. Concept / Approach: We use permutations because the arrangement of selected letters matters. We have 10 distinct letters and we are forming ordered arrangements of length 4. The number of such permutations is given by 10P4, which equals 10 * 9 * 8 * 7. This formula arises from counting choices step by step: first position has 10 choices, second has 9 remaining choices, and so on. Step-by-Step Solution: Step 1: Compute 10P4 for permutations of 10 distinct letters taken 4 at a time. Step 2: 10P4 = 10 * 9 * 8 * 7. Step 3: Calculate 10 * 9 = 90. Step 4: Then 90 * 8 = 720. Step 5: Finally, 720 * 7 = 5040. Step 6: Therefore, the number of different 4 letter words that can be formed is 5040. Verification / Alternative check: We can also check using the factorial formula for permutations: nPr = n! / (n - r)!. Here n = 10 and r = 4, so 10P4 = 10! / 6!. The value 10! equals 3628800 and 6! equals 720. Dividing 3628800 by 720 gives 5040, which matches our previous calculation. This consistency provides strong confirmation that we have counted correctly. Why Other Options Are Wrong: Values such as 4050, 3600 and 1200 do not equal 10P4. They typically arise from incomplete multiplication (for example, stopping at 10 * 9 * 8) or mixing combinations with permutations. For instance, 10C4 * 4! simplifies to 10P4, but if someone uses only 10C4, they will get a much smaller value. Only 5040 correctly represents the number of ordered 4 letter arrangements formed from 10 distinct letters without repetition. Common Pitfalls: A frequent error is to confuse this situation with combinations and use 10C4, which counts only unordered selections of letters. Another mistake is to allow repetition unintentionally, which would change the counting completely. Some learners also miscount the number of distinct letters in LOGARITHMS and treat it as 9 or 8 letters. Always verify the distinct letters and decide whether order matters before choosing between permutations and combinations. Final Answer: The number of different 4 letter words that can be formed from the letters of LOGARITHMS without repetition is 5040.

Question 18

When four fair dice are rolled simultaneously, in how many possible outcomes does at least one of the dice show the number 3 on its upper face?

Accepted Answer

Introduction / Context: This question tests the use of the complement principle in probability and counting. Counting the outcomes where at least one die shows a specific number is often easier by first counting all possible outcomes and then subtracting the outcomes where no die shows that number. Here, we focus on counting rather than probability, but the underlying logic is the same and very important for aptitude problems. Given Data / Assumptions: There are 4 fair dice rolled simultaneously. Each die has 6 faces numbered from 1 to 6. We are interested in outcomes where at least one die shows 3. Each die roll is independent of the others. Concept / Approach: The total number of outcomes when 4 dice are rolled is 6^4, since each die has 6 possible results. To find how many outcomes show at least one 3, it is easier to subtract from the total the number of outcomes where no die shows 3. If 3 is forbidden, each die has only 5 possible values (1, 2, 4, 5 or 6). Therefore, the number of outcomes with no 3 is 5^4. The required count is total outcomes minus outcomes with no 3. Step-by-Step Solution: Step 1: Compute the total number of outcomes when 4 dice are rolled. Step 2: Total outcomes = 6^4 = 6 * 6 * 6 * 6 = 1296. Step 3: Count outcomes where no die shows 3. Then each die has 5 options. Step 4: Outcomes with no 3 = 5^4 = 5 * 5 * 5 * 5 = 625. Step 5: Required outcomes with at least one 3 = total - no 3. Step 6: Compute 1296 - 625 = 671. Step 7: Therefore, there are 671 outcomes where at least one die shows the number 3. Verification / Alternative check: A direct method would try to count outcomes with exactly 1, 2, 3 or 4 dice showing 3 and sum those counts. Although possible, it is more tedious and prone to errors. For example, exactly one die showing 3 gives 4C1 * 5^3 outcomes, exactly two dice give 4C2 * 5^2, and so on. If you compute and sum these terms, you will again reach 671. This confirms that the complement method is both simpler and reliable. Why Other Options Are Wrong: The number 625 is the count of outcomes with no 3 at all, not the outcomes with at least one 3. The values 620 and 567 do not correspond to any standard combination or complement calculation in this context. The value 630 may arise from partial counting of only some cases. Only 671 agrees with the formula 6^4 - 5^4 and matches both the complement and expansion methods. Common Pitfalls: One common mistake is to misinterpret at least one as exactly one and count only one case. Another mistake is to forget to subtract from the total and instead add or multiply incorrectly. Some learners also miscalculate 6^4 or 5^4. Always break the problem into total outcomes and outcomes that do not satisfy the condition, then subtract. This complement idea appears frequently in exam questions involving dice, cards and other discrete random situations. Final Answer: The number of outcomes in which at least one die shows the number 3 is 671.

Question 19

In a plane there are 37 straight lines. Of these, 13 lines pass through point A and 11 lines pass through point B. No three lines (apart from those through A or B) pass through the same point, no line passes through both A and B, and no two lines ar…

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Introduction / Context: This problem combines general counting of intersections with corrections for special configurations where many lines meet at the same point. Normally, any two non parallel lines intersect in exactly one point, but here several lines share point A and point B. We must correctly count distinct intersection points while avoiding overcounting the intersections that occur at A and B. This is a common type of advanced combinatorics question in geometry. Given Data / Assumptions: Total number of lines in the plane = 37. 13 of these lines pass through point A. 11 of these lines pass through point B. No line passes through both A and B. No two lines are parallel. Except for the lines through A or B, no three lines pass through a single point. Concept / Approach: If no special configurations existed, any two lines would intersect at a unique point, and the total number of intersection points would be 37C2. However, at point A, the 13 lines intersect all together, so the naive count 37C2 includes many repeated counts of the same geometric point A. Similarly at point B. We first compute 37C2 and then subtract the extra counts at A and B. At point A, the number of pairs of lines passing through A is 13C2, all meeting at the single point A. This should contribute just 1 distinct intersection, not 13C2. The same logic holds for point B with 11C2 pairs. Step-by-Step Solution: Step 1: Compute the total number of pairwise intersections if all lines were in general position. Step 2: Total naive intersections = 37C2 = 37 * 36 / 2 = 666. Step 3: At point A, there are 13 lines, so the number of pairs among them is 13C2 = 13 * 12 / 2 = 78. Step 4: All these 78 pairs meet at the same point A, but we want to count that single point only once. Step 5: Hence, the overcount at A is 78 - 1 because we should have counted only one intersection point there. Step 6: Overcount at A = 78 - 1 = 77. Step 7: Similarly, at point B there are 11 lines, so pairs = 11C2 = 11 * 10 / 2 = 55. Step 8: All these 55 pairs meet at the same point B, but this should be counted as a single intersection point. Step 9: Overcount at B = 55 - 1 = 54. Step 10: Total overcount = 77 + 54 = 131. Step 11: Required number of distinct intersection points = 666 - 131 = 535. Verification / Alternative check: To cross-check, imagine replacing the clusters of lines at A and B with equivalent configurations where each pair intersection is distinct except one. For 13 lines at A, if they were in general position, we would have 78 distinct intersection points. But when they all pass through A, these 78 points collapse into one. This is exactly a loss of 77 distinct intersection points. Similarly, for 11 lines at B, 55 potential intersection points collapse into one, a loss of 54. Subtracting these losses from the naive count 666 gives 535. This conceptual view matches the detailed step-by-step subtraction and confirms the correctness. Why Other Options Are Wrong: The value 525 suggests subtracting an incorrect amount of overcount. Likewise 545 and 555 correspond to different incorrect corrections that do not match the actual multiplicities of intersections at A and B. If someone forgets to adjust for one of the special points or miscomputes 13C2 or 11C2, they may get one of these values. Only 535 correctly reflects the elimination of repeated counting at both special points A and B. Common Pitfalls: The most common error is to compute only 37C2 and stop, forgetting that many line pairs meet at exactly the same points A and B. Another pitfall is to subtract 13C2 and 11C2 without adding back 1 for each common point, effectively removing those intersections entirely instead of collapsing them into one. Some students also confuse conditions like no three lines concurrent and no two parallel. Careful reading and a clear understanding of overcounting corrections are essential. Final Answer: The total number of distinct intersection points determined by the 37 lines is 535.

Question 20

How many distinct necklaces can be formed using all 18 beads of different colours if arrangements that are the same under rotation and reflection are treated as identical?

Accepted Answer

Introduction / Context: This problem is a classic necklace counting question involving circular permutations where rotations and reflections of the same arrangement are considered identical. When beads are all different, the number of distinct circular arrangements is not simply n!. We must adjust for the fact that rotating or flipping the necklace does not produce a new pattern. Understanding this idea is very important in more advanced counting questions that involve symmetry. Given Data / Assumptions: Number of beads = 18. All 18 beads are of different colours. Beads are arranged to form a closed necklace (a circle). Arrangements that differ only by rotation are considered the same. Arrangements that differ only by reflection (flipping the necklace) are also considered the same. Concept / Approach: For n distinct objects arranged in a line, there are n! permutations. However, for a circular arrangement where rotations are considered identical, the number of distinct circular permutations reduces to (n - 1)!. This accounts for the fact that rotating the circle does not change the pattern. When reflections (mirror images) are also considered identical, each circular arrangement is counted twice (once for each orientation). Therefore, the number must be divided by 2 again, giving (n - 1)! / 2 distinct arrangements for a necklace with n distinct beads where both rotation and reflection equivalences are used. Step-by-Step Solution: Step 1: Let n = 18 since there are 18 distinct beads. Step 2: First consider circular permutations where rotations are identical but reflections are still considered different. Step 3: The count of such circular permutations is (n - 1)! = 17!. Step 4: Now incorporate the condition that reflections (mirror images) are not considered new arrangements. Step 5: For distinct beads, each circular arrangement has a mirror image, so each unique necklace pattern is counted twice in the 17! circular arrangements. Step 6: To correct this double counting, divide by 2. The number of distinct necklaces = 17! / 2. Step 7: Therefore, the required answer is 17!/2. Verification / Alternative check: We can confirm the formula by checking smaller values of n. For example, with 3 distinct beads, a line arrangement would give 3! = 6 permutations, but in a necklace with reflections, there is essentially only 1 distinct arrangement. Plugging n = 3 into the formula (n - 1)! / 2 gives 2! / 2 = 1, which matches. For 4 beads, (4 - 1)! / 2 = 3! / 2 = 3 distinct necklaces, which can also be verified by drawing. These checks support the general formula (n - 1)! / 2 when rotations and reflections are treated as equivalent. Why Other Options Are Wrong: The value 18! counts all linear permutations and ignores both circular symmetry and reflection symmetry. The value 18!/2 adjusts for only one type of symmetry and is still much too large. The option 17! corrects for rotation but not for reflection. The option 16!/2 has no logical basis in this context. Only 17!/2 correctly accounts for both rotational and reflection equivalence for a necklace with 18 distinct beads. Common Pitfalls: A very common mistake is to use n! directly, which is correct only for linear arrangements. Many students remember dividing by n to handle circular symmetry, obtaining (n - 1)!, but then forget to adjust for reflection when the problem specifically describes a necklace that can be flipped. Another pitfall is thinking that reflections are already included in the division by n, which is not true. To handle both rotation and reflection, always start from linear permutations n!, divide by n for rotation, and then divide by 2 for reflection, giving (n - 1)! / 2. Final Answer: The number of distinct necklaces that can be formed is 17!/2.

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